The Mean Value Theorem is the most important theorem in calculus! It allows us to infer information about a function from information about its derivative. Such as: a function whose derivative is zero must be a constant function.

1. Section 4.2
The Mean Value Theorem
V63.0121.006/016, Calculus I
March 25, 2010
Announcements
Please resubmit your exams to be my 4:00pm TODAY for
regrading (it’s only one problem, and scores will either go
up or stay the same)
Quiz 3 in recitation Friday, April 2. Covers §§2.6–3.5.
Contact me if you have religious conflicts.
. . . . . .

2. Announcements
Please resubmit your exams to be my 4:00pm TODAY for
regrading (it’s only one problem, and scores will either go
up or stay the same)
Quiz 3 in recitation Friday, April 2. Covers §§2.6–3.5.
Contact me if you have religious conflicts.
. . . . . .

3. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
. . . . . .

4. Flowchart for placing extrema
Thanks to Fermat
Suppose f is a continuous function on the closed, bounded
interval [a, b], and c is a global maximum point.
.
. . c is a
start
local max
. . .
Is c an Is f diff’ble f is not
n
.o n
.o
endpoint? at c? diff at c
y
. es y
. es
. .
c = a or
f′ (c) = 0
c = b
. . . . . .

5. The Closed Interval Method
This means to find the maximum value of f on [a, b], we need
to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points x where either f′ (x) = 0 or
f is not differentiable at x.
The points with the largest function value are the global
maximum points
The points with the smallest or most negative function
value are the global minimum points.
. . . . . .

6. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
. . . . . .

7. Heuristic Motivation for Rolle's Theorem
If you bike up a hill, then back down, at some point your
elevation was stationary.
.
.
Image credit: SpringSun . . . . . .

8. Mathematical Statement of Rolle's Theorem
Theorem (Rolle’s Theorem)
Let f be continuous on
[a, b] and differentiable
on (a, b). Suppose
f(a) = f(b). Then there
exists a point c in (a, b)
such that f′ (c) = 0.
. . .
a
. b
.
. . . . . .

9. Mathematical Statement of Rolle's Theorem
Theorem (Rolle’s Theorem)
c
..
Let f be continuous on
[a, b] and differentiable
on (a, b). Suppose
f(a) = f(b). Then there
exists a point c in (a, b)
such that f′ (c) = 0.
. . .
a
. b
.
. . . . . .

10. Flowchart proof of Rolle's Theorem
.
. . endpoints
Let c be
. Let d be
. .
are max
the max pt the min pt
and min
.
. . f is
is c. an is d. an. .
y
. es y
. es constant
endpoint? endpoint?
on [a, b]
n
.o n
.o
.
.
. ′
. ′ f′ (x) .≡ 0
f (c) .= 0 f (d) .= 0
on (a, b)
. . . . . .

11. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
. . . . . .

12. Heuristic Motivation for The Mean Value Theorem
If you drive between points A and B, at some time your
speedometer reading was the same as your average speed
over the drive.
.
. . . . . .

13. The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on
[a, b] and differentiable
on (a, b). Then there
exists a point c in (a, b)
such that .
b
.
f(b) − f(a)
= f′ (c). . .
b−a a
.
. . . . . .

14. The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on
[a, b] and differentiable
on (a, b). Then there
exists a point c in (a, b)
such that .
b
.
f(b) − f(a)
= f′ (c). . .
b−a a
.
. . . . . .

15. The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on c
.
[a, b] and differentiable
on (a, b). Then there
exists a point c in (a, b)
such that .
b
.
f(b) − f(a)
= f′ (c). . .
b−a a
.
. . . . . .

16. Rolle vs. MVT
f(b) − f(a)
f′ (c) = 0 = f′ (c)
b−a
c
.. c
..
.
b
.
. . . . .
a
. b
. a
.
. . . . . .

17. Rolle vs. MVT
f(b) − f(a)
f′ (c) = 0 = f′ (c)
b−a
c
.. c
..
.
b
.
. . . . .
a
. b
. a
.
If the x-axis is skewed the pictures look the same.
. . . . . .

18. Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation
f(b) − f(a)
y − f(a) = (x − a)
b−a
. . . . . .

19. Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation
f(b) − f(a)
y − f(a) = (x − a)
b−a
Apply Rolle’s Theorem to the function
f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
. . . . . .

20. Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation
f(b) − f(a)
y − f(a) = (x − a)
b−a
Apply Rolle’s Theorem to the function
f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
Then g is continuous on [a, b] and differentiable on (a, b) since f
is.
. . . . . .

21. Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation
f(b) − f(a)
y − f(a) = (x − a)
b−a
Apply Rolle’s Theorem to the function
f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
Then g is continuous on [a, b] and differentiable on (a, b) since f
is. Also g(a) = 0 and g(b) = 0 (check both)
. . . . . .

22. Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation
f(b) − f(a)
y − f(a) = (x − a)
b−a
Apply Rolle’s Theorem to the function
f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
Then g is continuous on [a, b] and differentiable on (a, b) since f
is. Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’s
Theorem there exists a point c in (a, b) such that
f(b) − f(a)
0 = g′ (c) = f′ (c) − .
b−a
. . . . . .

23. Using the MVT to count solutions
Example
Show that there is a unique solution to the equation
x3 − x = 100 in the interval [4, 5].
. . . . . .

24. Using the MVT to count solutions
Example
Show that there is a unique solution to the equation
x3 − x = 100 in the interval [4, 5].
Solution
By the Intermediate Value Theorem, the function
f(x) = x3 − x must take the value 100 at some point on c in
(4, 5).
. . . . . .

25. Using the MVT to count solutions
Example
Show that there is a unique solution to the equation
x3 − x = 100 in the interval [4, 5].
Solution
By the Intermediate Value Theorem, the function
f(x) = x3 − x must take the value 100 at some point on c in
(4, 5).
If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
then somewhere between them would be a point c3
between them with f′ (c3 ) = 0.
. . . . . .

26. Using the MVT to count solutions
Example
Show that there is a unique solution to the equation
x3 − x = 100 in the interval [4, 5].
Solution
By the Intermediate Value Theorem, the function
f(x) = x3 − x must take the value 100 at some point on c in
(4, 5).
If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
then somewhere between them would be a point c3
between them with f′ (c3 ) = 0.
However, f′ (x) = 3x2 − 1, which is positive all along (4, 5).
So this is impossible.
. . . . . .

27. Example
We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|.
. . . . . .

28. Example
We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|.
Solution
Apply the MVT to the function f(t) = sin t on [0, x]. We get
sin x − sin 0
= cos(c)
x−0
for some c in (0, x). Since |cos(c)| ≤ 1, we get
sin x
≤ 1 =⇒ |sin x| ≤ |x|
x
. . . . . .

29. Example
Let f be a differentiable function with f(1) = 3 and f′ (x) < 2 for
all x in [0, 5]. Could f(4) ≥ 9?
. . . . . .

30. Example
Let f be a differentiable function with f(1) = 3 and f′ (x) < 2 for
all x in [0, 5]. Could f(4) ≥ 9?
Solution
By MVT y
. . 4, 9) .
(
f(4) − f(1)
= f′ (c) < 2 . 4, f(4))
(
.
4−1
for some c in (1, 4). Therefore
f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9. .
. 1, 3)
(
So no, it is impossible that
f(4) ≥ 9.
. x
.
. . . . . .

31. Question
A driver travels along the New Jersey Turnpike using E-ZPass.
The system takes note of the time and place the driver enters
and exits the Turnpike. A week after his trip, the driver gets a
speeding ticket in the mail. Which of the following best
describes the situation?
(a) E-ZPass cannot prove that the driver was speeding
(b) E-ZPass can prove that the driver was speeding
(c) The driver’s actual maximum speed exceeds his ticketed
speed
(d) Both (b) and (c).
Be prepared to justify your answer.
. . . . . .

32. Question
A driver travels along the New Jersey Turnpike using E-ZPass.
The system takes note of the time and place the driver enters
and exits the Turnpike. A week after his trip, the driver gets a
speeding ticket in the mail. Which of the following best
describes the situation?
(a) E-ZPass cannot prove that the driver was speeding
(b) E-ZPass can prove that the driver was speeding
(c) The driver’s actual maximum speed exceeds his ticketed
speed
(d) Both (b) and (c).
Be prepared to justify your answer.
. . . . . .

33. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
. . . . . .

34. Fact
If f is constant on (a, b), then f′ (x) = 0 on (a, b).
. . . . . .

35. Fact
If f is constant on (a, b), then f′ (x) = 0 on (a, b).
The limit of difference quotients must be 0
The tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
Implied by the power rule since c = cx0
. . . . . .

36. Fact
If f is constant on (a, b), then f′ (x) = 0 on (a, b).
The limit of difference quotients must be 0
The tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
Implied by the power rule since c = cx0
Question
If f′ (x) = 0 is f necessarily a constant function?
. . . . . .

37. Fact
If f is constant on (a, b), then f′ (x) = 0 on (a, b).
The limit of difference quotients must be 0
The tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
Implied by the power rule since c = cx0
Question
If f′ (x) = 0 is f necessarily a constant function?
It seems true
But so far no theorem (that we have proven) uses
information about the derivative of a function to determine
information about the function itself
. . . . . .

38. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b).
. . . . . .

39. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
. . . . . .

40. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous
on [x, y] and differentiable on (x, y). By MVT there exists a point
z in (x, y) such that
f(y) − f(x)
= f′ (z) = 0.
y−x
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
. . . . . .

41. Theorem
Suppose f and g are two differentiable functions on (a, b) with
f′ = g′ . Then f and g differ by a constant. That is, there exists a
constant C such that f(x) = g(x) + C.
. . . . . .

42. Theorem
Suppose f and g are two differentiable functions on (a, b) with
f′ = g′ . Then f and g differ by a constant. That is, there exists a
constant C such that f(x) = g(x) + C.
Proof.
Let h(x) = f(x) − g(x)
Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
So h(x) = C, a constant
This means f(x) − g(x) = C on (a, b)
. . . . . .

43. MVT and differentiability
Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0
Is f differentiable at 0?
. . . . . .

44. MVT and differentiability
Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0
Is f differentiable at 0?
Solution (from the definition)
We have
f(x) − f(0) −x
lim = lim = −1
x→0 − x−0 x→0 − x
f(x) − f(0) x2
lim+ = lim+ = lim+ x = 0
x→0 x−0 x→0 x x→0
Since these limits disagree, f is not differentiable at 0.
. . . . . .

45. MVT and differentiability
Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0
Is f differentiable at 0?
Solution (Sort of)
If x < 0, then f′ (x) = −1. If x > 0, then f′ (x) = 2x. Since
lim f′ (x) = 0 and lim f′ (x) = −1,
x→0+ x→0−
the limit lim f′ (x) does not exist and so f is not differentiable at 0.
x→0
. . . . . .

46. Why only “sort of"?
This solution is valid but .′ (x)
f
less direct. y
. f
.(x)
We seem to be using
the following fact: If
lim f′ (x) does not exist,
x→a
then f is not
. x
.
differentiable at a.
equivalently: If f is .
differentiable at a, then
lim f′ (x) exists.
x→a
But this “fact” is not
true!
. . . . . .

47. Differentiable with discontinuous derivative
It is possible for a function f to be differentiable at a even if
lim f′ (x) does not exist.
x→a
Example
{
′ x2 sin(1/x) if x ̸= 0
Let f (x) = . Then when x ̸= 0,
0 if x = 0
f′ (x) = 2x sin(1/x)+x2 cos(1/x)(−1/x2 ) = 2x sin(1/x)−cos(1/x),
which has no limit at 0. However,
f(x) − f(0) x2 sin(1/x)
f′ (0) = lim = lim = lim x sin(1/x) = 0
x→0 x−0 x→0 x x→0
So f′ (0) = 0. Hence f is differentiable for all x, but f′ is not
continuous at 0!
. . . . . .

48. Differentiability FAIL
f
.(x) .′ (x)
f
. x
. . x
.
This function is differentiable But the derivative is not
at 0. continuous at 0!
. . . . . .

49. MVT to the rescue
Lemma
Suppose f is continuous on [a, b] and lim+ f′ (x) = m. Then
x→a
f(x) − f(a)
lim = m.
x→a+ x−a
. . . . . .

50. MVT to the rescue
Lemma
Suppose f is continuous on [a, b] and lim+ f′ (x) = m. Then
x→a
f(x) − f(a)
lim = m.
x→a+ x−a
Proof.
Choose x near a and greater than a. Then
f(x) − f(a)
= f′ (cx )
x−a
for some cx where a < cx < x. As x → a, cx → a as well, so:
f(x) − f(a)
lim = lim+ f′ (cx ) = lim+ f′ (x) = m.
x→a+ x−a x→a x→a
. . . . . .

51. Theorem
Suppose
lim f′ (x) = m1 and lim+ f′ (x) = m2
x→a− x→a
If m1 = m2 , then f is differentiable at a. If m1 ̸= m2 , then f is not
differentiable at a.
. . . . . .

52. Theorem
Suppose
lim f′ (x) = m1 and lim+ f′ (x) = m2
x→a− x→a
If m1 = m2 , then f is differentiable at a. If m1 ̸= m2 , then f is not
differentiable at a.
Proof.
We know by the lemma that
f(x) − f(a)
lim = lim f′ (x)
x→a− x−a x→a−
f(x) − f(a)
lim+ = lim+ f′ (x)
x→a x−a x→a
The two-sided limit exists if (and only if) the two right-hand
sides agree.
. . . . . .

53. What have we learned today?
Rolle’s Theorem: under suitable conditions, functions must
have critical points.
Mean Value Theorem: under suitable conditions, functions
must have an instantaneous rate of change equal to the
average rate of change.
A function whose derivative is identically zero on an
interval must be constant on that interval.
E-ZPass is kinder than we realized.
. . . . . .